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# There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h

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Re: There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h [#permalink]
To get differrent colours, let see the differrent possibiities:
Red and green = 3C1 * 5C1 = 15
Red and blue = 3C1 * 6C1 = 18
Green and blue = 5C1 * 6C1 = 30
No. Of ways of picking two differrent colours = 15+18+30 = 63.

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Re: There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h [#permalink]
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Mseemab wrote:
There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In how many ways can we pick two books without replacement such that both books have different colors?

A. 28
B. 63
C. 25
D. 52
E. 60

Experts please explain, what can be the possible ways to approach such questions?

The total number of ways to select two books from 14 is 14C2 = 14! / (12! x 2!) = (14 x 13)/2 = 7 x 13 = 91.

The number of ways to select 2 red books is 3C2 = 3.

The number of ways to select 2 green books is 5C2 = 5! / (3! x 2!) = (5 x 4)/2 = 10.

The number of ways to select 2 blue books is 6C2 = 6! / (4! x 2!) = (6 x 5)/2 = 15.

So the number of ways to select books of different colors is:

91 - 10 - 15 - 3 = 91 - 28 = 63

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Re: There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h [#permalink]
Total ways to select:
1 red + 1 any other = 3 x 11 = 33
1 green + 1 any other = 5 x 9 = 45
1 blue + 1 any other = 6 x 8 = 48

Total ways = 126

Divide by 2 (since blue/green and green/blue is the same) , therefore 63

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Re: There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h [#permalink]
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Re: There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h [#permalink]
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